<< problem 313 - Sliding game Firecracker - problem 317 >>

# Problem 315: Digital root clocks

Sam and Max are asked to transform two digital clocks into two "digital root" clocks.
A digital root clock is a digital clock that calculates digital roots step by step.

When a clock is fed a number, it will show it and then it will start the calculation, showing all the intermediate values until it gets to the result.
For example, if the clock is fed the number 137, it will show: "137" → "11" → "2" and then it will go black, waiting for the next number.

Every digital number consists of some light segments: three horizontal (top, middle, bottom) and four vertical (top-left, top-right, bottom-left, bottom-right).
Number "1" is made of vertical top-right and bottom-right, number "4" is made by middle horizontal and vertical top-left, top-right and bottom-right. Number "8" lights them all.

The clocks consume energy only when segments are turned on/off.
To turn on a "2" will cost 5 transitions, while a "7" will cost only 4 transitions.

Sam and Max built two different clocks.

Sam's clock is fed e.g. number 137: the clock shows "137", then the panel is turned off, then the next number ("11") is turned on, then the panel is turned off again and finally the last number ("2") is turned on and, after some time, off.
For the example, with number 137, Sam's clock requires:
"137": (2 + 5 + 4) * 2 = 22 transitions ("137" on/off).
"11": (2 + 2) * 2 = 8 transitions ("11" on/off).
"2": (5) * 2 = 10 transitions ("2" on/off).
For a grand total of 40 transitions.

Max's clock works differently. Instead of turning off the whole panel, it is smart enough to turn off only those segments that won't be needed for the next number.
For number 137, Max's clock requires:
"137": 2 + 5 + 4 = 11 transitions ("137" on)
7 transitions (to turn off the segments that are not needed for number "11").
"11": 0 transitions (number "11" is already turned on correctly)
3 transitions (to turn off the first "1" and the bottom part of the second "1"; the top part is common with number "2").
"2": 4 transitions (to turn on the remaining segments in order to get a "2")
5 transitions (to turn off number "2").
For a grand total of 30 transitions.

Of course, Max's clock consumes less power than Sam's one. The two clocks are fed all the prime numbers between A = 10^7 and B = 2 * 10^7.

Find the difference between the total number of transitions needed by Sam's clock and that needed by Max's one.

# My Algorithm

I encode all lit segments as bitmasks in Segments. Counting the set bits (popcnt) is equivalent to counting the lit segments.
There are at most 8 digits, hence a 64 bit integer represents all segments of a number.

Sam's algorithm can be found in sam: switching on and off all segments "costs" 2x popcnt(getSegments(x)).

Max's algorithm (see max) is a bit more complicated: I have to compare all lit segments of the previous number to the lit segments of the current number.
When I XOR both bitmasks then I get a new bitmasks with all segments that change (either have to be turned off or turnd on).
Counting the set bits in the XORed value tells me the number of transitions.
Before the first and after the last number all lit segments transition from and to the "empty" state.

Both routines work recursively because I hoped to improve the execution time by caching / memoizing results:
After the first step the digitSum is below 70. However, the simple memoization in sam didn't help much
and therefore I didn't even attempt to code something similiar in max.

## Note

popcnt was named after the super-fast CPU instruction which counts bits.
My G++ compiler shows the result after 0.12 seconds when that intrinsic can be used but needs 0.17 seconds with my own alternative code
(which still relies on tricky bit manipulation, see bits.stephan-brumme.com).

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):
Note: Enter the begin and the end of the range to be analyzed.

This is equivalent to
echo "137 137" | ./315

Output:

(please click 'Go !')

Note: the original problem's input 10000000 20000000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <iostream>
#include <vector>

// represent digits as bitmasks, lit segments are 1
const unsigned char Segments[10] =
{
// assume the following numbering scheme for the seven segments:
//  000
// 1   2
// 1   2
//  333
// 4   5
// 4   5
//  666
(1 << 0) | (1 << 1) | (1 << 2) | (0 << 3) | (1 << 4) | (1 << 5) | (1 << 6), // digit 0 = 0x77
(0 << 0) | (0 << 1) | (1 << 2) | (0 << 3) | (0 << 4) | (1 << 5) | (0 << 6), // digit 1 = 0x24
(1 << 0) | (0 << 1) | (1 << 2) | (1 << 3) | (1 << 4) | (0 << 5) | (1 << 6), // digit 2 = 0x5d
(1 << 0) | (0 << 1) | (1 << 2) | (1 << 3) | (0 << 4) | (1 << 5) | (1 << 6), // digit 3 = 0x6d
(0 << 0) | (1 << 1) | (1 << 2) | (1 << 3) | (0 << 4) | (1 << 5) | (0 << 6), // digit 4 = 0x2e
(1 << 0) | (1 << 1) | (0 << 2) | (1 << 3) | (0 << 4) | (1 << 5) | (1 << 6), // digit 5 = 0x6b
(1 << 0) | (1 << 1) | (0 << 2) | (1 << 3) | (1 << 4) | (1 << 5) | (1 << 6), // digit 6 = 0x3b
(1 << 0) | (1 << 1) | (1 << 2) | (0 << 3) | (0 << 4) | (1 << 5) | (0 << 6), // digit 7 = 0x27
(1 << 0) | (1 << 1) | (1 << 2) | (1 << 3) | (1 << 4) | (1 << 5) | (1 << 6), // digit 8 = 0x7f
(1 << 0) | (1 << 1) | (1 << 2) | (1 << 3) | (0 << 4) | (1 << 5) | (1 << 6)  // digit 9 = 0x6f
};

// ---------- standard prime sieve from my toolbox ----------

// odd prime numbers are marked as "true" in a bitvector
std::vector<bool> sieve;

// return true, if x is a prime number
bool isPrime(unsigned int x)
{
// handle even numbers
if ((x & 1) == 0)
return x == 2;

// lookup for odd numbers
return sieve[x >> 1];
}

// find all prime numbers from 2 to size
void fillSieve(unsigned int size)
{
// store only odd numbers
const unsigned int half = (size >> 1) + 1;

// allocate memory
sieve.resize(half, true);
// 1 is not a prime number
sieve[0] = false;

// process all relevant prime factors
for (unsigned int i = 1; 2 * i*i < half; i++)
// do we have a prime factor ?
if (sieve[i])
{
// mark all its multiples as false
unsigned int current = 3 * i + 1;
while (current < half)
{
sieve[current] = false;
current += 2 * i + 1;
}
}
}

// ---------- problem solution ----------

// return sum of all digits
unsigned int digitSum(unsigned int x)
{
unsigned int result = 0;
while (x > 0)
{
result += x % 10;
x      /= 10;
}
return result;
}

// convert number to its bitmasks, 8 bits per digit => x must not exceed 99999999
unsigned long long getSegments(unsigned int x)
{
unsigned long long result = 0;
unsigned int shift = 0;
while (x > 0)
{
unsigned long long current = Segments[x % 10];
result |= current << shift;
x      /= 10;
shift  += 8;
}
return result;
}

// count bits which are set to 1
unsigned int popcnt(unsigned long long x)
{
#ifdef __GNUC__
return __builtin_popcount(x);
#endif

unsigned int result = 0;
while (x > 0)
{
// clear right-most bit
x &= x - 1;
result++;
}
return result;
}

// process all steps with strategy of Sam, return number of transitions
unsigned int sam(unsigned int x)
{
// memoize (runs quite fast without, too)
static std::vector<unsigned int> cache(100, 0);
if (x < cache.size() && cache[x] > 0)
return cache[x];

// display was clear
auto segments = getSegments(x);
// switch (intermediate) number on and off
auto result   = 2 * popcnt(segments);

// need further iterations ?
if (x > 9)
result += sam(digitSum(x));

if (x < cache.size())
cache[x] = result;
return result;
}

// process all steps with strategy of Max, return number of transitions
unsigned int max(unsigned int x, unsigned long long previousSegments = 0)
{
auto segments    = getSegments(x);
auto transitions = segments ^ previousSegments;
auto result      = popcnt(transitions);

if (x > 9)
// morph to next number
result += max(digitSum(x), segments);
else
// fade out
result += popcnt(segments);

return result;
}

int main()
{
// search range
unsigned int from = 10000000;
unsigned int to   = 20000000;
std::cin >> from >> to;

// for live input only
if (from > to)
return 1;

// generate prime sieve
fillSieve(to);

// count all digits of Sam and Max
unsigned long long sumSam = 0;
unsigned long long sumMax = 0;
for (auto i = (from | 1); i <= to; i += 2) // odd numbers only
{
if (!isPrime(i))
continue;

sumSam += sam(i);
sumMax += max(i);
}

// Sam >= Max, display the difference
std::cout << sumSam - sumMax << std::endl;
return 0;
}


This solution contains 27 empty lines, 39 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.13 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 3 MByte.

(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

July 29, 2017 submitted solution
July 29, 2017 added comments

# Difficulty

Project Euler ranks this problem at 20% (out of 100%).

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

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